Solve Using the Quadratic Formula 4x^2-4x=8

$4 x^{2} - 4 x = 8$

Step 1

Subtract $8$ from both sides of the equation.

$4 x^{2} - 4 x - 8 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 4$ , $b = - 4$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (4 \cdot - 8)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 8}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 8$ .

Multiply $- 4$ by $4$ .

$x = \frac{4 \pm \sqrt{16 - 16 \cdot - 8}}{2 \cdot 4}$

Multiply $- 16$ by $- 8$ .

$x = \frac{4 \pm \sqrt{16 + 128}}{2 \cdot 4}$

Add $16$ and $128$ .

$x = \frac{4 \pm \sqrt{144}}{2 \cdot 4}$

Rewrite $144$ as $12^{2}$ .

$x = \frac{4 \pm \sqrt{12^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{4 \pm 12}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{4 \pm 12}{8}$

Simplify $\frac{4 \pm 12}{8}$ .

$x = \frac{1 \pm 3}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 2, - 1$